Understanding $\Bbb Z_3[X] /(x^3+x^2+x+1) $

Question

In $\Bbb Z_3[x]$ let $f(x)=x^3+x^2+x+1$ and let $A=\Bbb Z_3[x]/(f) $ be the quotient ring. I'm asked to find the cardinality of $A $ and whether it is a field.

For the cardinality, I should find that $A $ is in bijection with the set of polynomials with degree smaller than $3$ (and in $\Bbb Z_3[X]$), hence $|A|=3^3$.The thing is, I can't really visualise $A$. I know its elements are the classes of the remainders of Euclidean division by $f$, but how do I get these?

For the other question, the given answer is that $f(x)=(x^2+1)(x+1) $ is reducible in $\Bbb Z_3[X]$, thus $A$ is not a field. Why does this work? Is this a general property of quotient rings of polynomials, i.e. if $a\in A[X]$ is reducible then $A[X]/(a)$ is not a field?

Answer 1

$\frac {\mathbb Z_3[x]}{x^3+x^2+x+1}$ gives us a set of polynomials in $\mathbb Z_3[x]$ of degree 2 (or less).

This can be thought of as a 3 dimensional vector space. And there are $3^3$ elements.

Since $(x + 1)(x^2 +1) = 0$ in this ring, it is not a field. In a field $ab = 0$ if and only if $a = 0$ or $b = 0$

Answer 2

I'm guessing by $\mathbb{Z}_3$ you mean $\mathbb{Z}/3\mathbb{Z}$ and not the $3$-adic integers.

To answer your second question, yes there is a general property of this sort. Namely, a quotient ring $R/I$ is a field if and only if $I$ is a maximal ideal. In your case, $I$ is the principal ideal generated by $f$, so $I$ is maximal iff $f$ is irreducible (prove this yourself!).

Your first question seems more conceptual. A way to "visualize" $A$ could be by using the chinese remainder theorem to write it as the product of a field with 9 elements and a field with 3 elements via the decomposition of $f$ you gave. Have a look at the related questions if you're still confused.